\(n_{Al_2O_3}=\dfrac{153}{102}=1,5\left(mol\right)\)
ta có: \(n_{Al}=2n_{Al_2O_3}=2\times1,5=3\left(mol\right)\)
\(\Rightarrow m_{Al}=3\times27=81\left(g\right)\)
Ta có: \(n_O=3n_{Al_2O_3}=3\times1,5=4,5\left(mol\right)\)
\(\Rightarrow m_O=4,5\times16=72\left(g\right)\)